$f(x, y) = (-xy^2, yx^2)$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-y^2 - 2xy, x^2 + 2xy)$ (Choice B) B $(-y^2, 2xy)$ (Choice C) C $(-2y, 2x)$ (Choice D) D $(-2xy, x^2)$
Answer: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y) = (f_0(x, y), f_1(x, y)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $y$, we'll treat $x$ as if it were a constant. Therefore, $f_y = (-2xy, x^2)$.